AMC 10 Step by Step

2011 AMC 10A Problem 15

P11-15~3 minPrint

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?

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Key insight

Miles per gallon for the whole trip is total miles divided by total gallons. Only the miles after the first 40 use any gas.

Solution

"Averaged 55 miles per gallon" means: total miles divided by total gallons of gas equals 5555. So we need both totals.

1. Let the trip be xx miles long.
2. The first 4040 miles use no gas. The rest of the trip is x−40x - 40 miles, and each of those miles uses 0.020.02 gallons. So the total gas is 0.02(x−40)0.02(x - 40) gallons.
3. Set total miles divided by total gallons equal to 5555:
x0.02(x−40)=55.\frac{x}{0.02(x - 40)} = 55.
4. Multiply both sides by the bottom: x=55×0.02×(x−40)x = 55 \times 0.02 \times (x - 40). Since 55×0.02=1.155 \times 0.02 = 1.1, this is
x=1.1(x−40)=1.1x−44.x = 1.1(x - 40) = 1.1x - 44.
5. Subtract xx from both sides: 0=0.1x−440 = 0.1x - 44, so 0.1x=440.1x = 44 and x=440x = 440.

Check: the gas part is 440−40=400440 - 40 = 400 miles, using 400×0.02=8400 \times 0.02 = 8 gallons. And 440÷8=55440 \div 8 = 55. ✓

The answer is (C) 440\boxed{\textbf{(C)}\ 440}.

Why this works

Miles per gallon for a whole trip is one big ratio: all the miles over all the gallons. On gas alone the car gets 1÷0.02=501 \div 0.02 = 50 miles per gallon. The free battery miles add distance without using gas, which is why the trip average (5555) is higher than 5050. Once you write both totals correctly, the equation is a simple one.

Alternative approach

Test the answer choices, since each one takes only a moment.

  • 240240 miles: 200200 gas miles use 44 gallons, and 240÷4=60240 \div 4 = 60. Too high.
  • 440440 miles: 400400 gas miles use 88 gallons, and 440÷8=55440 \div 8 = 55. ✓
  • 640640 miles: 600600 gas miles use 1212 gallons, and 640÷12≈53.3640 \div 12 \approx 53.3. Too low.

Only 440440 gives exactly 5555.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let the trip be xx miles. The gasoline portion is x−40x - 40 miles, consuming 0.02(x−40)0.02(x - 40) gallons. The battery portion uses no gasoline.

The trip's average is total miles divided by total gallons:
x0.02(x−40)=55.\frac{x}{0.02(x - 40)} = 55.
Multiply out: x=55⋅0.02 (x−40)=1.1(x−40)=1.1x−44x = 55 \cdot 0.02\,(x - 40) = 1.1(x - 40) = 1.1x - 44. So 0.1x=440.1x = 44 and x=440x = 440.

Check: 400400 gasoline miles use 88 gallons, and 440/8=55440/8 = 55.

The answer is (C) 440\boxed{\textbf{(C)}\ 440}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.